Search any question & find its solution
Question:
Answered & Verified by Expert
When a $100 \mathrm{Kg}$ mass is suspended from wire of length 1 meter and and area $1 \mathrm{~cm}^2$, it length increases.
Find the elastic potential energy stored in the wire
Options:
Find the elastic potential energy stored in the wire
Solution:
2434 Upvotes
Verified Answer
The correct answers are:
$5 \times 10^{-2} \mathrm{~J}$
$\checkmark$ Answer
Given $\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^2$ and Young Modulus of the material of the wire $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
Stress $=$ Load/Area
$=\frac{m g}{A}=\frac{100 \times 10}{10^4}=10^7 \mathrm{~N} / \mathrm{m}^2$
Now
Young Modulus=Stress/Strain
So strain $=$ Stress/Young Modulus
$=5 \times 10^{-5}$
Now we know that
Strain $=$ Increase in length/Initial length
$\Rightarrow$ Increase in Length=strain * Initial length
$=5 \times 10^{-5} \mathrm{~m}$
Elastic Potential energy
$\begin{aligned} & =\frac{1}{2} \text { stress } \times \text { strain } \times \text { Volume } \\ & =5 \times 10^{-2} \mathrm{~J}\end{aligned}$
Given $\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^2$ and Young Modulus of the material of the wire $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
Stress $=$ Load/Area
$=\frac{m g}{A}=\frac{100 \times 10}{10^4}=10^7 \mathrm{~N} / \mathrm{m}^2$
Now
Young Modulus=Stress/Strain
So strain $=$ Stress/Young Modulus
$=5 \times 10^{-5}$
Now we know that
Strain $=$ Increase in length/Initial length
$\Rightarrow$ Increase in Length=strain * Initial length
$=5 \times 10^{-5} \mathrm{~m}$
Elastic Potential energy
$\begin{aligned} & =\frac{1}{2} \text { stress } \times \text { strain } \times \text { Volume } \\ & =5 \times 10^{-2} \mathrm{~J}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.